example 1\ngraph the function $y = -\\frac{1}{2}\\cos x$.

Explanation:

Step1: Determine the amplitude

The amplitude of (y = A\cos x) is (|A|). For (y =-\frac{1}{2}\cos x), (A =-\frac{1}{2}), so the amplitude (|A|=\frac{1}{2}).

Step2: Analyze the reflection

The negative sign in (y =-\frac{1}{2}\cos x) reflects the graph of (y=\cos x) about the (x -)axis.

Step3: Plot key points

For (y = \cos x), key points are ((0,1)), ((\frac{\pi}{2},0)), ((\pi,- 1)), ((\frac{3\pi}{2},0)), ((2\pi,1)). For (y =-\frac{1}{2}\cos x):

• When (x = 0), (y=-\frac{1}{2}\cos(0)=-\frac{1}{2}(1)=-\frac{1}{2})
• When (x=\frac{\pi}{2}), (y =-\frac{1}{2}\cos(\frac{\pi}{2})=0)
• When (x=\pi), (y=-\frac{1}{2}\cos(\pi)=-\frac{1}{2}(-1)=\frac{1}{2})
• When (x = \frac{3\pi}{2}), (y=-\frac{1}{2}\cos(\frac{3\pi}{2})=0)
• When (x=2\pi), (y=-\frac{1}{2}\cos(2\pi)=-\frac{1}{2}(1)=-\frac{1}{2})

Step4: Sketch the graph

Connect the key points ((0,-\frac{1}{2})), ((\frac{\pi}{2},0)), ((\pi,\frac{1}{2})), ((\frac{3\pi}{2},0)), ((2\pi,-\frac{1}{2})) with a smooth curve. The graph has a period of (2\pi) (same as (y = \cos x) since there is no horizontal - scaling, (B = 1) in (y=A\cos(Bx))), amplitude (\frac{1}{2}), and is reflected about the (x -)axis.

Answer:

The graph of (y =-\frac{1}{2}\cos x) has amplitude (\frac{1}{2}), is a reflection of (y = \cos x) about the (x -)axis, and has key points ((0,-\frac{1}{2})), ((\frac{\pi}{2},0)), ((\pi,\frac{1}{2})), ((\frac{3\pi}{2},0)), ((2\pi,-\frac{1}{2})) which are connected with a smooth cosine - like curve with period (2\pi).